Problem: You have found the following ages (in years) of 6 turtles. Those turtles were randomly selected from the 43 turtles at your local zoo: $ 47,\enspace 24,\enspace 44,\enspace 27,\enspace 28,\enspace 101$ Based on your sample, what is the average age of the turtles? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 43 turtles, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{47 + 24 + 44 + 27 + 28 + 101}{{6}} = {45.2\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {3.24} + {449.44} + {1.44} + {331.24} + {295.84} + {3113.64}} {{6 - 1}} $ {s^2} = \dfrac{{4194.84}}{{5}} = {838.97\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{838.97\text{ years}^2}} = {29\text{ years}} $ We can estimate that the average turtle at the zoo is 45.2 years old. There is also a standard deviation of 29 years.